AQA A-Level Chemistry AS Paper 1, June 2023: Question 8

9 marks · Medium difficulty · State/Explain/Numerical

Calculate equilibrium amounts, write the expression for Kc, explain the effect of temperature on Kc, and calculate the equilibrium concentration of a reactant.

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Question

Question 08 focuses on the reversible aqueous reaction A(aq) + 2B(aq) ⇌ C(aq) with an enthalpy change of -32 kJ mol⁻¹. In 08.1, candidates deduce equilibrium amounts of A and B given initial amounts of 0.60 mol each and 0.28 mol of C formed. In 08.2, write an expression for Kc. In 08.3, predict and explain the effect on Kc when temperature is decreased. In 08.4, calculate equilibrium concentration of B given [A] = 0.48 mol dm⁻³, [C] = 0.62 mol dm⁻³, and Kc = 7.8 mol⁻² dm⁶ to appropriate significant figures.
Question text

08 This question is about the equilibrium mixture formed when A and B react.

A(aq) + 2 B(aq) ⇌ C(aq) ΔH= –32 kJ mol–1

08.1 A solution containing 0.60 mol of A is added to a solution containing 0.60 mol of B.

The amount of C formed at equilibrium is 0.28 mol

Deduce the amounts, in moles, of A and B in this mixture at equilibrium.

[2 marks]

Amount of A mol

Amount of B mol

08.2 Give an expression for the equilibrium constant (Kc) for this reaction.

[1 mark]

Kc

08.3 The temperature of the equilibrium mixture is decreased.

Predict the effect, if any, on the value of Kc

Give a reason for your prediction.

[3 marks]

Prediction

Reason

08.4 In another mixture at equilibrium

[A] = 0.48 mol dm–3

[C] = 0.62 mol dm–3

For this reaction, the equilibrium constant K = 7.8 mol–2 dm6

c

Calculate [B] at equilibrium.

Give your answer to the appropriate number of significant figures.

[3 marks]

[B] mol dm–3

Mark scheme

Show the mark scheme Mark scheme for Question 08. 08.1 gives Amount of A = 0.32 mol and Amount of B = 0.04 mol (2 marks). 08.2 gives Kc = [C] / ([A][B]²) (1 mark). 08.3 awards 3 marks for: M1 greater/increases, M2 equilibrium shifts in exothermic direction, M3 to increase temperature / oppose temperature decrease. 08.4 awards 3 marks: M1 rearranging to [B]² = [C]/([A]Kc) or [B] = sqrt(0.62/(0.48 x 7.8)), M2 evaluation to 0.406938, M3 rounding correctly to 2 significant figures as 0.41.

Question Marking guidance Additional Comments/Guidelines Mark

Amount of A = 0.32 mol

08.1

(2 x AO2)

Amount of B = 0.04 mol

[C] 1

08.2 Kc = 2

[A] [B] (AO1)

M1 greater / increases

08.3 M2 the equilibrium shifts in the exothermic direction

(3 x AO3)

M3 to increase the temperature / to oppose the temperature decrease

2 [C] 2 [C] 0.62 M1 – allow consequential to incorrect 08.2

M1 [B] = or [B] = 0.16559829 or [B] = � or [B] = �

[A] Kc [A] Kc 0.48 𝑥𝑥 7.8

M2 – evaluation of [B] from M1

08.4 M2 [B] = 0.406938

M3 – M2 given to 2sf (3 x AO2)

M3 [B] = 0.41 Answer to 2sf

How to answer it

Equilibrium Quantities, Kc Expressions, and Le Chatelier's Principle

WHAT THIS QUESTION TESTS

This question assesses your mastery of homogeneous equilibria and equilibrium constants in aqueous systems:

  • Applying reacting stoichiometry (ICE table concept) to determine equilibrium moles from initial amounts.
  • Writing a correct homogeneous equilibrium constant (Kc) expression with proper concentration notation.
  • Predicting and explaining the effect of temperature on the value of Kc using Le Chatelier's principle and enthalpy change (ΔH).
  • Algebraically rearranging a Kc expression to solve for an unknown equilibrium concentration with appropriate significant figures.
Reaction: A(aq) + 2B(aq) ⇌ C(aq)   |   ΔH = −32 kJ mol⁻¹
PART 08.1 • 2 MARKS

Deducing Equilibrium Amounts in Moles

Calculate the amounts of A and B at equilibrium given 0.60 mol initial reactants and 0.28 mol C formed

📐 Step-by-Step Stoichiometric Calculation

Species A 2B C
Initial 0.60 mol 0.60 mol 0.00 mol
Change −0.28 mol −2 × (0.28) = −0.56 mol +0.28 mol
Equilibrium 0.32 mol 0.04 mol 0.28 mol

1 mol of A forms 1 mol of C → moles of A reacted = 0.28 mol.
2 mol of B form 1 mol of C → moles of B reacted = 2 × 0.28 = 0.56 mol.

✅ Correct Answers

Amount of A = 0.32 mol [1 mark]

Amount of B = 0.04 mol [1 mark]

❌ Common Errors

  • Ignoring the 1:2 stoichiometry: Many students subtract 0.28 from 0.60 for B, incorrectly writing B = 0.32 mol.
  • Subtracting from the equilibrium amount of C instead of the initial amount of reactant.
Mark Breakdown: 1 mark for 0.32 mol A; 1 mark for 0.04 mol B. (AO2)
PART 08.2 • 1 MARK

Equilibrium Constant Expression

State the mathematical expression for Kc

✅ Correct Answer

Kc = [C] / ([A][B]²)

🧠 Exam Technique & Notation

  • Always use square brackets [ ] to denote concentration in mol dm⁻³. Regular parentheses ( ) are strictly penalised.
  • Powers must equal the balancing numbers: B has a coefficient of 2, so [B] must be squared.
  • No state symbols are required inside the Kc expression.
Mark Breakdown: 1 mark for the fully correct expression with square brackets and [B]² in the denominator. (AO1)
PART 08.3 • 3 MARKS

Effect of Temperature on Kc

Predict and explain the effect on Kc when the temperature decreases

✅ Correct Answer

Prediction: Greater / Increases [1 mark]

Reason:

  • The equilibrium shifts in the exothermic direction [1 mark]
  • to increase the temperature / oppose the temperature decrease [1 mark]

💡 Key Knowledge: Why Kc Changes

  • Temperature is the ONLY factor that changes the value of Kc. (Pressure, concentration, and catalysts do not change Kc).
  • ΔH = −32 kJ mol⁻¹ means the forward reaction is exothermic (releases heat).
  • Lowering temperature favours the heat-producing (exothermic / forward) reaction.
  • More products [C] and fewer reactants [A] and [B] at equilibrium mean the ratio [C]/([A][B]²) increases, so Kc increases.

❌ Common Errors & Examiner Commentary

  • Stating only that the equilibrium "shifts right" without explicitly stating that the forward direction is exothermic.
  • Forgetting the Le Chatelier link: stating what happens (shifts exothermic) without stating why (to oppose the drop in temperature / raise temperature).
  • Confusing equilibrium position with rate: saying "reaction slows down so Kc decreases". Rate of reaction has no bearing on equilibrium constant.
Mark Breakdown: M1: Prediction: greater/increases; M2: Equilibrium shifts in the exothermic direction; M3: To oppose the temperature decrease / to increase the temperature. (3 × AO3)
PART 08.4 • 3 MARKS

Equilibrium Concentration Calculation

Calculate [B] at equilibrium given [A] = 0.48, [C] = 0.62, and Kc = 7.8 mol⁻² dm⁶

📐 Step-by-Step Calculation

  1. Rearrange the Kc expression for [B]²:
    Kc = [C] / ([A][B]²)
    [B]² = [C] / ([A] × Kc) [M1]
  2. Substitute the given values:
    [B]² = 0.62 / (0.48 × 7.8)
    [B]² = 0.62 / 3.744 = 0.165598...
  3. Take the square root to find [B]:
    [B] = √(0.165598...) = 0.406938... mol dm⁻³ [M2]
  4. Round to appropriate significant figures:
    Given data: [A] = 0.48 (2 s.f.), [C] = 0.62 (2 s.f.), Kc = 7.8 (2 s.f.).
    [B] = 0.41 mol dm⁻³ (2 s.f.) [M3]

✅ Correct Answer

[B] = 0.41 mol dm⁻³

Must be exactly 2 significant figures to match the provided data.

❌ Calculation Traps

  • Forgetting to square root: Leaving the answer as 0.17 mol dm⁻³ (which is [B]², not [B]).
  • Incorrect rearrangement: Writing [B]² = ([C] × Kc) / [A] or [B]² = ([A] × Kc) / [C].
  • Significant figure penalty: Writing 0.407 or 0.4 loses the 3rd mark. Always look at the data values in the stem: all are 2 s.f.
  • Premature rounding: Rounding 0.165598 to 0.16 before taking the root yields 0.40 instead of 0.41. Keep full calculator precision throughout!
Mark Breakdown: M1: Rearrangement [B]² = [C] / ([A]Kc) or [B]² = 0.166; M2: Correct square root evaluation (0.407); M3: Final answer given strictly to 2 significant figures (0.41). (3 × AO2)

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.